Solution:

Let's verify and calculate the correct answer step-by-step:

Given Data:

• Capacitances:
  $C_1 = 2 \, \mu\text{F}, C_2 = 3 \, \mu\text{F}, C_3 = 5 \, \mu\text{F}$ 
• Maximum voltages:
  $V_1 = 3 \, \text{V}, V_2 = 2 \, \text{V}, V_3 = 1 \, \text{V}$ 

Step 1: Maximum charge each capacitor can store:

$$Q_1 = C_1 \cdot V_1 = 2 \cdot 3 = 6 \, \mu\text{C}$$

$$Q_2 = C_2 \cdot V_2 = 3 \cdot 2 = 6 \, \mu\text{C}$$

$$Q_3 = C_3 \cdot V_3 = 5 \cdot 1 = 5 \, \mu\text{C}$$

The capacitor with the minimum charge capacity limits the system. Here,

$Q_{\text{max}} = 5 \, \mu\text{C}$

, dictated by

$C_3$

.

Step 2: Voltage distribution across each capacitor:

In series, charge

$Q$

is the same on all capacitors, and the voltage across each capacitor is:

$$V_1 = \frac{Q}{C_1}, \quad V_2 = \frac{Q}{C_2}, \quad V_3 = \frac{Q}{C_3}$$

Total voltage across the series combination:

$$V_{\text{total}} = V_1 + V_2 + V_3$$

Substitute

$Q = 5 \, \mu\text{C}$

:

$$V_1 = \frac{5}{2} = 2.5 \, \text{V}, \quad V_2 = \frac{5}{3} \approx 1.67 \, \text{V}, \quad V_3 = \frac{5}{5} = 1 \, \text{V}$$

Step 3: Total voltage:

$$V_{\text{total}} = V_1 + V_2 + V_3 = 2.5 + 1.67 + 1 = \frac{15}{6} + \frac{10}{6} + \frac{6}{6} = \frac{31}{6} \, \text{V}.$$

Final Answer:

The maximum voltage the series combination can withstand is:

$$\boxed{\frac{31}{6} \, \text{V}} \approx 5.17 \, \text{V}.$$